QuestionApril 16, 2026

Ammonia is produced from the reaction of nitrogen and hydrogen gases. What volume of hydrogen at STP is required to completely react and produce 25 of ammonia? N_(2)(g)+3H_(2)(g)-2NH_(3)(g) A 33 B 1.1 L C 491 D 251

Ammonia is produced from the reaction of nitrogen and hydrogen gases. What volume of hydrogen at STP is required to completely react and produce 25 of ammonia? N_(2)(g)+3H_(2)(g)-2NH_(3)(g) A 33 B 1.1 L C 491 D 251
Ammonia is produced from the reaction of nitrogen and hydrogen gases. What volume of hydrogen at STP is required to completely react and produce 25 of ammonia?
N_(2)(g)+3H_(2)(g)-2NH_(3)(g)
A 33
B 1.1 L
C 491
D 251

Solution
4.5(219 votes)

Answer

49.1 L Explanation 1. Determine molar mass and moles of ammonia Molar mass of NH_3 = 14 + (3 \times 1) = 17 \ \text{g/mol}. Moles of NH_3: n = \frac{25}{17} \approx 1.47 \ \text{mol}. 2. Use stoichiometric ratio From N_2 + 3H_2 \rightarrow 2NH_3: 2 \ \text{mol} \ NH_3 needs 3 \ \text{mol} \ H_2. So 1.47 \ \text{mol} \ NH_3 needs: 1.47 \times \frac{3}{2} \approx 2.205 \ \text{mol} \ H_2. 3. Convert moles to volume at STP At STP, **V = n \times 22.4 \ \text{L}**. V_{H_2} = 2.205 \times 22.4 \approx 49.4 \ \text{L}.

Explanation

1. Determine molar mass and moles of ammonia <br /> Molar mass of $NH_3$ = $14 + (3 \times 1) = 17 \ \text{g/mol}$. <br />Moles of $NH_3$: $n = \frac{25}{17} \approx 1.47 \ \text{mol}$. <br /><br />2. Use stoichiometric ratio <br /> From $N_2 + 3H_2 \rightarrow 2NH_3$: <br />$2 \ \text{mol} \ NH_3$ needs $3 \ \text{mol} \ H_2$. <br />So $1.47 \ \text{mol} \ NH_3$ needs: $1.47 \times \frac{3}{2} \approx 2.205 \ \text{mol} \ H_2$. <br /><br />3. Convert moles to volume at STP <br /> At STP, **$V = n \times 22.4 \ \text{L}$**. <br />$V_{H_2} = 2.205 \times 22.4 \approx 49.4 \ \text{L}$.
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